# Thread: Linear operators and Differential Equations

1. ## Linear operators and Differential Equations

Hello,

I have missed all my lectures on the above topic due to illness. Does anyone know of a good website with solved excercise sheets to help me learn material on this topic, in particular, I am looking for helpful notes for homework questions on this website http://www.maths.qmul.ac.uk/~cchu/MTH6122/lode092.pdf

Many Thanks

2. $= \int_{0} ^{1} ( \int_{0} ^{1} f(y)K(x,y)dy )g(x)dx = \int_{0} ^{1} ( \int_{0} ^{1} g(x)f(y)K(x,y)dy )dx$ Using Fubini, this last one is equal to $\int_{0} ^{1} ( \int_{0} ^{1} g(x)f(y)K(x,y)dx )dy = \int_{0} ^{1} ( \int_{0} ^{1} g(x)K(x,y)dx )f(y)dy = $ So $L ^* =L$ ie. $L$ is self-adjoint.

Now suppose $f$ is an eigenfunction of $L$ with eigenvalue $\lambda$ with $K$ as defined in the sheet then:

$\lambda f(x) = \int_{0} ^{1} f(y)K(x,y)dy = \int_{0} ^{x} f(y)(1-x)ydy + \int_{x} ^{1} f(y)(1-y)xdy$ $= (1-x) \int_{0} ^{x} yf(y)dy + x\int_{x} ^{1} (1-y)f(y)dy$. In using this last expression for $\lambda f(x)$ it's clear that $f(0)=0=f(1)$ (as long as $\lambda \neq 0$, a difficulty that'll dissapear in the next step).

Now, assuming $f$ is twice differentiable (only continuity will not be enough) we have:

$\lambda f'(x)=-\int_{0} ^{x} yf(y)dy +(1-x)xf(x) + \int_{x}^{1} (1-y)f(y)dy - x(1-x)f(x)$ $= -\int_{0}^{x} yf(y)dy +\int_{x}^{1} (1-y)f(y)dy$

$\lambda f''(x)=-xf(x) - (1-x)f(x)=-f(x)$ and so $\lambda f'' +f=0$. Now just notice that we never used $\lambda \neq 0$ so if $\lambda =0$ we just have $f=0$ which satisfies the boundary value problem trivially.

3. You are a life saver Jose! I will donate some money to this great site tonight.